Exercise 5.26

I’m going to show this for vector-valued functions $f=(f_1,\dots ,f_k)$ mapping $[a,b]$ into $R^k$ such that $f(a)=0$ and such that $|f^{\prime }(x)|\le A|f(x)|$, since this is needed for Exercise 28.

If $A=0$, then $f^{\prime }=0$, so $f=f(a)=0$ by Theorem 5.11(b). So assume that $A>0$. Following the hint, let $\delta >0$ be small enough so that $\delta <1∕A$ and $x_0=a+\delta \le b$. For $i=1,\dots ,k$, let $M_{i,0}=\sup <!--\; FUNCTION\; APPLICATION\; -->|f_i(x)|$, $M_{i,1}=\sup <!--\; FUNCTION\; APPLICATION\; -->|\underset{i}{\overset{\prime }{f}}(x)|$ for $a\le x\le x_0$. By Theorem 5.19, for any $x\in (a,x_0)$ and for every $i=1,\dots ,k$ there is a $t_i\in (a,x)$ such that

Since $\mathit{A\delta }<1$, this can only happen if $M_{i,0}=0$, so $f_i(x)=0$ in $[a,a+\delta ]$ for each $i=1,\dots ,k$, that is, $f=0$ in $[a,a+\delta ]$. Repeating this argument with $a+\delta$ replacing $a$, we get $f(x)=0$ in $[a,a+2\delta ]$. After about $(b-a)∕\delta$ steps, we’ve shown that $f(x)=0$ in $[a,b]$.